Question

prove that (a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)=2(a^3b^3c^3-3abc)

Answer 1

hopefully these will help uh...if it is plus sign..nd if it is in multiple thn sorry i can't slove tht...

Answer 2

Answer: Proved.

Step-by-step explanation:  We are given to prove the following:

$(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)=2(a^3+b^3+c^3-3abc).$

We will be using the following cubic formula in the proof:

$(a+b)^3=a^3+b^3+3a^2b+3ab^2.$

We have

$L.H.S.\\\\=(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)\\\\=a^3+b^3+3a^2b+3ab^2+b^3+c^3+3b^2c+3bc^2+c^3+a^3+3c^2a+3ca^2-3(ab+ac+b^2+bc)(c+a)\\\\=2a^3+2b^3+2c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2-3abc-3ac^2-3b^2c-3bc^2-3a^2b-3a^2c-3ab^2-3abc\\\\=2a^3+2b^3+2c^3-6abc\\\\=2(a^3+b^3+c^3-3abc)\\\\=R.H.S.$

Hence proved.